Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
 
int main() {
	int n;
	cin >> n;
	vector<long double> p(n+1);
	for(int i = 1; i<=n; i++){
		cin >> p[i];
	}
	vector<long double> dp(n+1);
	dp[0] = 1;
 
	for(int i = 1; i<= n ; i++){
		for(int j = i; j >=0 ; j--){
			dp[j] = (j?dp[j-1]*p[i]:0) + (dp[j]*(1-p[i]));
		}
	}
 
	long double res = 0;
	for(int i = (n/2)+1; i <=n ; i++){
		res += dp[i];
	}
	cout<<setprecision(10)<<res<<endl;
 
	return 0;
}